12 ms·
It's not even a transfer since you can't control the state of the measured atom. It just collapses into one state which happens to also affect the other atom. S
by collaborative 4y ago
It's not even a transfer since you can't control the state of the measured atom. It just collapses into one state which happens to also affect the other atom. So you could never choose to transfer actual data
- nerdponx 4y agoMy understanding is that it's more like putting two copies of a letter into an envelope, mailing one copy halfway around the world, and then opening the other copy. You now know what they both contain because you know they contain the same thing. Is this at least somewhat directionally correct? I know that quantum physics principles are very difficult to define without going into the actual math.
- affgrff 4y agoFunny how similar our comments are, I swear I believe I have not conciously read your comment before writing mine.
- eloff 4y agoThat was Einstein's view, but it was disproved in the seventies by experimental measurements of Bell's inequality. I'm just a computer guy, so someone else can chime in with a more detailed explanation. It's also easy to find more info on Google.
- rhinokungfoo 4y agoYour example is akin to the Hidden variable theory (in the example the hidden variable will be the fact the two letters are copies). Bell's inequality tests have ruled out the hidden variable theory. Veritasium has a good video explaining Bell's inequality - https://www.youtube.com/watch?v=ZuvK-od647c https://www.youtube.com/watch?v=ZuvK-od647c
- zeroonetwothree 4y agoIt only rules out local hidden variable theories that assume statistical independence. Of course many scientists don’t like violating locality or statistical independence but there’s no philosophical reason to a priori discard them.
- affgrff 4y agoMy interpretation is that is not better for communication than putting the same random number in two envelopes, sending them somewhere and then opening both at the same time.
- cortic 4y agoIf A and B are entangled, and you observe B, would there be any way of knowing B was observed from observing A? If yes, that's binary communication. If no, then how do we know B changed at all?
- cortic 4y agocorrection; If no, then how do we know A changed at all (when we observe B)?
- l33tman 4y agoYou don't, until you meet up and compare notes.
- cortic 4y agoOkay, so how do we know that A wasn't like that all along? edit; Been trying to read up on this, i think i get it; Its a random outcome whether B has been looked at or not, but when you check them later, they are opposite pairs randomly distributed.
- l33tman 4y agoYes exactly. Nothing "changes" at the other end (although, there has been arguments regarding the Wigner's Friend thought experiments about this... see Guerin 2021) Just be careful when reading up on this about a subtlety here - as long as both A and B measure the same property, say "spin along Y axis", a measurement they will always both measure the opposite of, the experiment will seem just like the "unopened envelopes" popular description of entanglement which is not really magical. It's annoying that this particular description keeps being made.. The interesting things happen when you start mixing up the measurements so A and B measure different, but correlated, properties. In the context of Bell's tests, they will not measure in the exact same basis (like vertical polarization), one of them will rotate the basis slightly, say 30 or 60 degrees. And the resulting correlations between A and B in this scenario can't be explained by hidden variables ("unopened envelopes describing the result of all possible measurements") but is predicted correctly by quantum mechanics.